(i) Consider the line element describing the interior of a star,

$$d s^{2}=e^{2 \alpha(r)} d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)-e^{2 \gamma(r)} d t^{2},$$

defined for $0 \leq r \leq r_{0}$ by

$$e^{-2 \alpha(r)}=1-A r^{2}$$

and

$$e^{\gamma(r)}=\frac{3}{2} e^{-\alpha_{0}}-\frac{1}{2} e^{-\alpha(r)}$$

Here $A=2 M / r_{0}^{3}, M$ is the mass of the star, and $\alpha_{0}$ is defined to be $\alpha\left(r_{0}\right)$.

The star is made of a perfect fluid with energy-momentum tensor

$$T_{a b}=(p+\rho) u_{a} u_{b}+p g_{a b} .$$

Here $u^{a}$ is the 4 -velocity of the fluid which is at rest, the density $\rho$ is constant throughout the star $\left(0 \leq r \leq r_{0}\right)$ and the pressure $p=p(r)$ depends only on the radial coordinate. Write down the Einstein field equations and show that (in geometrical units with $G=c=1$ ) they may equivalently be written as

$$R_{a b}=8 \pi(p+\rho) u_{a} u_{b}+4 \pi(p-\rho) g_{a b} .$$

(ii) Using the formulae below, or otherwise, show that for $0 \leq r \leq r_{0}$ one has

$$\rho=\frac{3 A}{8 \pi}, \quad p(r)=\frac{3 A}{8 \pi}\left(\frac{e^{-\alpha(r)}-e^{-\alpha_{0}}}{3 e^{-\alpha_{0}}-e^{-\alpha(r)}}\right)$$

[The non-zero components of the Ricci tensor are:

$$\begin{gathered} R_{11}=-\gamma^{\prime \prime}+\alpha^{\prime} \gamma^{\prime}-\gamma^{\prime 2}+\frac{2 \alpha^{\prime}}{r}, \quad R_{22}=e^{-2 \alpha}\left[\left(\alpha^{\prime}-\gamma^{\prime}\right) r-1\right]+1 \\ R_{33}=\sin ^{2} \theta R_{22}, \quad R_{44}=e^{2 \gamma-2 \alpha}\left[\gamma^{\prime \prime}-\alpha^{\prime} \gamma^{\prime}+\gamma^{\prime 2}+\frac{2 \gamma^{\prime}}{r}\right] \end{gathered}$$

Note that

$$\alpha^{\prime}=A r e^{2 \alpha}, \quad \gamma^{\prime}=\frac{1}{2} A r e^{\alpha-\gamma}, \quad \gamma^{\prime \prime}=\frac{1}{2} A e^{\alpha-\gamma}+\frac{1}{2} A^{2} r^{2} e^{3 \alpha-\gamma}-\frac{1}{4} A^{2} r^{2} e^{2 \alpha-2 \gamma} \text { ] ] }$$